Divide the following complex numbers. $ \dfrac{6-14i}{2-2i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+2i}$ $ \dfrac{6-14i}{2-2i} = \dfrac{6-14i}{2-2i} \cdot \dfrac{{2+2i}}{{2+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(6-14i) \cdot (2+2i)} {(2-2i) \cdot (2+2i)} = \dfrac{(6-14i) \cdot (2+2i)} {2^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(6-14i) \cdot (2+2i)} {(2)^2 - (-2i)^2} = $ $ \dfrac{(6-14i) \cdot (2+2i)} {4 + 4} = $ $ \dfrac{(6-14i) \cdot (2+2i)} {8} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({6-14i}) \cdot ({2+2i})} {8} = $ $ \dfrac{{6} \cdot {2} + {-14} \cdot {2 i} + {6} \cdot {2 i} + {-14} \cdot {2 i^2}} {8} $ Evaluate each product of two numbers. $ \dfrac{12 - 28i + 12i - 28 i^2} {8} $ Finally, simplify the fraction. $ \dfrac{12 - 28i + 12i + 28} {8} = \dfrac{40 - 16i} {8} = 5-2i $